The Toronto problem

Topology and its Applications 162 (2014) 76–90

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Topology and its Applications www.elsevier.com/locate/topol

The Toronto problem William Rea Brian Tulane University, Department of Mathematics, 6823 St. Charles Avenue, New Orleans, LA 70118, United States

a r t i c l e

i n f o

Article history: Received 3 December 2012 Received in revised form 13 November 2013 Accepted 20 November 2013 MSC: primary 54G99 secondary 54A35, 54G12

a b s t r a c t A Toronto space is a topological space that is homeomorphic to every one of its full-cardinality subspaces, and the Toronto problem asks whether every Hausdorff Toronto space of size ℵ1 is discrete. We examine compactness properties, convergence properties, and separation properties of non-discrete Hausdorff Toronto spaces of size ℵ1 , and we classify the non-T1 Toronto spaces of any infinite size. © 2013 Elsevier B.V. All rights reserved.

Keywords: Toronto space Toronto problem

1. Introduction A Toronto space is a topological space X such that Y is homeomorphic to X whenever Y ⊆ X and |Y | = |X|. For a given infinite cardinal κ, there are a handful of easy examples of Toronto spaces of size κ: the discrete and indiscrete topologies, the co-finite topology, and more generally, the topology consisting of sets with complement smaller than λ for some λ  κ. We will outline a few more examples in Section 6, but none of them, aside from the discrete space, is Hausdorff. It remains an open (and seemingly very difficult!) problem whether it is consistent with ZFC to have an infinite Hausdorff non-discrete Toronto space. This problem has a straightforward solution for cardinality ℵ0 : every countable Hausdorff Toronto space is discrete. We will go a step further in Corollary 6.2 below and show that there are, up to homeomorphism, exactly five countable Toronto spaces. It is worth noting, however, that there is a variant of the definition of a Toronto space that turns out to be very interesting for countable spaces; these have been explored extensively in [4] and will be mentioned again in Section 4. The question of classifying the Toronto spaces of higher cardinality turns out to be much harder, and generalizes what has come to be known as the Toronto problem. The original Toronto problem, as posed in [6], asks whether every Hausdorff Toronto space of size ℵ1 is discrete. This question is our primary focus E-mail address: [email protected]. 0166-8641/$ – see front matter © 2013 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.topol.2013.11.009

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here. We will have frequent occasion to refer to a non-discrete Hausdorff Toronto space of size ℵ1 ; we will refer to such spaces as non-discrete HATS (Hausdorff Aleph-one Toronto Spaces). We write X ≈ Y whenever X is homeomorphic to Y . Borrowing notation from model theory and set theory, if S ⊆ X is some defined subset of X, say S = {x ∈ X: φ(x, X, p)}, and if Y ⊆ X is some subset of X of full cardinality, then Y ≈ X and we will denote the set in Y that corresponds to S, namely {y ∈ Y : φ(y, Y, p)}, by S Y . Equivalently, if S ⊆ X and h : X → Y is a homeomorphism, we write S Y for h[S] whenever h[S] does not depend on our choice of homeomorphism. 2. Some basic folklore In this section and the next two we will collect together some basic results about Toronto spaces. Many of these results are already known but their proofs have not been published elsewhere: instead they have become part of the “folklore” surrounding the Toronto problem. With the exception of Propositions 2.6 and 4.6 (which are due to the author), the results in Sections 2 and 4 are a group effort of Toronto-based topologists, most notably Alan Dow, Juris Stepr¯ ans, and Bill Weiss. The material in Section 3 is original except for Theorem 3.1, which is a rediscovery of an unpublished result of Ken Kunen. For any topological space X, we can define simultaneously by transfinite induction two sequences of subsets of X: X 0 = X,   IαX = x: x is isolated in X α , X α+1 = X α \ IαX ,  X β for limit α. Xα = β∈α

The sequence X α : α ∈ Ord is a decreasing sequence of closed subsets of X. Since X is not a proper class, there will be some β such that, for all γ  β, X γ = X β ; the least such β is called the Cantor–Bendixson rank of X and X β is the perfect kernel of X. This X β is the largest subset of X with no isolated points. If the perfect kernel of X is empty, i.e., if every subspace of X contains an isolated point, then X is said  to be scattered. Equivalently, X is scattered if and only if X = α<β IαX . Where no confusion can result, we will simply write Iα for IαX . If X is scattered, we will refer to the Cantor–Bendixson rank of X as the height of X, denoted ht(X). Also, the width of X, denoted wd(X), is the supremum of the cardinalities of the Iα . X is called thin–tall if wd(X) < ht(X). It is a nontrivial problem to show that thin–tall Hausdorff spaces exist, but ZFC examples have been constructed; see, for instance, [5]. If X is scattered and x ∈ X, the rank of x in X, denoted rkX (x), is the least α such that x ∈ / X α+1 , or, equivalently, the rank of x is the unique α such that x ∈ Iα . When no confusion will result, we will write rk(x) for rkX (x). Lemma 2.1. If Y is a subspace of a scattered space X then Y is scattered and, for every x ∈ Y , rkY (x)  rkX (x). If Y is open in X then rkY (x) = rkX (x) for every x ∈ Y . Continuing the analogy with model theory and set theory, Lemma 2.1 is something of an absoluteness result: it tells us how rkX (x) and rkY (x) relate to each other when Y ⊆ X and, in particular, that rank is “absolute” for open subsets. Proposition 2.2. Let X be a non-discrete HATS. Then X is a thin–tall scattered space with ht(X) = ω1 , and wd(X) = ω.

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Proof. Let x and y be distinct points of X. There are disjoint open sets Ux and Uy separating these points, so at least one of them, say Ux , must have uncountable complement. Then (X \ Ux ) ∪ {x} is uncountable and has an isolated point. Hence X contains an isolated point. But X cannot have only a finite set I0 of isolated points, because then X \ I0 ≈ X, would have no isolated points. X cannot have an uncountable set I0 of isolated points, since then I0 ≈ X is discrete. Thus |I0 | = ℵ0 . Since X 1 = X \ I0 ≈ X, we have 1 |I1X | = |I0X | = ℵ0 , and X 2 = X \ (I0 ∪ I1 ) ≈ X. In general, it follows by transfinite induction that if α < ω1 then X α ≈ X and |Iα | = ℵ0 : the successor step of this induction is the same as for the case α = 1 shown  above, and when α is a limit ordinal, X α = X \ β<α Iβ is uncountable and hence homeomorphic to X.  Now, as the Iα are disjoint, α<ω1 Iα is an uncountable subset of X, and hence homeomorphic to X. Since  α<ω1 Iα is a scattered space of width ω and height ω1 , the result is proved. 2 Proposition 2.3. If X is a non-discrete HATS then X is hereditarily separable but not Lindelöf. Proof. X is separable because I0 is dense in X. Every subset of X is either homeomorphic to X or countable, so X is hereditarily separable. X is not Lindelöf because the open cover {X \ X α : α < ω1 } has no countable subcover. 2 In fact, any non-discrete HATS X is hereditarily separable and anti-Lindelöf, i.e., every Lindelöf subset of X is countable. If in addition X is regular, then X is an S-space. The existence of S-spaces is known to be independent of ZFC, so it is consistent that there are no regular non-discrete HATS. The next theorem shows that, even without the requirement of regularity, it is consistent with ZFC that all HATS are discrete. Theorem 2.4. Suppose there is a non-discrete HATS. Then 2ℵ0 = 2ℵ1 . Proof. Suppose X is a non-discrete HATS and let   S = f : f is a homeomorphism X → Y, Y ⊆ X, and I0X ⊆ Y . Let Y be any uncountable subset of X that contains I0X . Clearly, there are 2ℵ1 such sets. Each such Y is homeomorphic to X under some map f ∈ S, so |S|  2ℵ1 . Each f ∈ S is a homeomorphism and therefore must map isolated points to isolated points and nonf [X] isolated points to non-isolated points: i.e., f [I0X ] = I0 = I0X . Thus f acts as a permutation on I0 . Furthermore, since I0 is dense in X and X is Hausdorff, f is determined uniquely by its action on I0 . In other words, every f ∈ S is determined uniquely by a permutation of I0 . As there are 2ℵ0 such permutations, |S|  2ℵ0 . 2 Corollary 2.5. It is consistent with ZFC that every HATS is discrete. Although our focus here is on Toronto spaces of size ℵ1 , the following observation is worthy of note: Proposition 2.6. Suppose that 2λ < 2κ whenever λ and κ are cardinals with λ < κ (this is true, for example, under GCH). Then every Hausdorff Toronto space (of any size) is discrete. Proof. Let κ be an infinite cardinal and let X be a non-discrete Hausdorff Toronto space of size κ. The case κ = ℵ0 will be covered in Corollary 6.2, so assume κ is uncountable. As in the proof of Proposition 2.2, X must have a nonempty set I0X of isolated points, and this set must have size λ for some infinite λ < κ. Working by induction just as in the proof of Proposition 2.2, one may show that X is a thin–tall scattered space with width λ and height κ. Following the proof of Theorem 2.4, one may then show that if such a space exists then 2λ = 2κ . 2

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3. Compactness properties In the last section we saw that non-discrete HATS do not necessarily exist. In the next three sections we will explore the question of what a non-discrete HATS would look like if it did exist. We begin by looking at some compactness properties. For what follows, take X to be a non-discrete HATS. Theorem 3.1 (Kunen). ω + 1 cannot be embedded in X. Proof. Suppose that f : ω + 1 → X is an embedding. Let   α = sup rk(x): x ∈ f [ω + 1]   and consider Y = f [ω + 1] ∪ X α+1 . f [ω + 1] is open in Y because f [ω + 1] = Y ∩ βα IβX and βα IβX is open in X. f [ω + 1] is closed in Y because it is compact and Y is Hausdorff. Thus f [ω + 1] is a clopen subset of Y that is homeomorphic to ω + 1, and it is clear that f (n) ∈ I0Y for each n while f (ω) ∈ I1Y . Since X ≈ Y , we have shown that some x ∈ I1 has a clopen neighborhood homeomorphic to ω + 1. Next we will show that every x ∈ I1 has a clopen neighborhood homeomorphic to ω + 1. Let us say that P is the property of having such a neighborhood. We have shown that at least one x ∈ I1 has property P . For each x ∈ I1 with property P , choose a clopen neighborhood Ux of x such that Ux ≈ ω + 1. Let A = {x ∈ I1 : x has property P } = ∅. Suppose A is countable. Then Y = X \ A ≈ X. Let x ∈ I1Y have a clopen neighborhood U ≈ ω + 1 (some such point must exist since X ≈ Y ). I1Y = Y ∩I1X by Lemma 2.1, so x ∈ I1X . Viewed as a neighborhood in X rather than Y , U is open (because Y is open in X) and closed (because U is compact and X is Hausdorff). Thus x has property P and is in A, a contradiction. Thus A is uncountable; set Y = I0 ∪ A. Clearly I1Y = A. If x ∈ A = I1Y and Ux is a clopen neighborhood of x in X that is homeomorphic to ω + 1, then Ux ∩ Y is a clopen neighborhood of x in Y that is homeomorphic to ω + 1. Thus every point of I1Y has property P in Y . Since Y ≈ X, every point of I1X has property P . 1 Let x ∈ I2 . X ≈ X 1 , so x ∈ I2X = I1X has a clopen neighborhood (in X 1 ) that is homeomorphic to ω + 1. 1 Let U  be a neighborhood of x in X 1 such that U  is clopen in X 1 , U  ∩ I1X = {x}, and U  ∩ X 1 ≈ ω + 1. Let U be some open set in X such that U ∩ X 1 = U  . Note that U ∩ X 2 = {x}. Write U  = U ∩X 1 = {xn : n ∈ ω}∪{x}; note that the map sending n to xn and ω to x is an embedding of ω + 1. For each n, let Un be a clopen neighborhood (in X) of xn that isolates xn in I1 and is homeomorphic  to ω + 1. Set Vn = Un \ m
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Proof. It suffices to show that if some limit x ∈ X has a countable neighborhood basis then ω + 1 can be embedded in X. As X is Hausdorff and not discrete, this is obvious. 2 Corollary 3.3. X is not metrizable. Corollary 3.4. X is not a GO space (recall that a GO space, or Generalized Order space, is any topology that is homeomorphic to a subspace of a linearly ordered space). Proof. X is locally countable, meaning that every x ∈ X has a countable open neighborhood, e.g. X \ X rk(x)+1 . By a theorem of van Dalen and Wattel (see [3] for a streamlined proof), a T1 space is a GO space if and only if it has a subbasis consisting of a union of two nests of open sets. Using this characterization, it is straightforward to show that any locally countable GO space must also be first countable. In fact, one may show that any locally countable space is first countable if its topology has a subbasis that is a union of countably many nests. Together with Corollary 3.2, this proves the stronger result that the topology on X is not generated by a countable union of nests. 2 Corollary 3.5. X is anti-compact, i.e., every compact subset of X is finite. Proof. Suppose K is an infinite compact subset of X. Note that K is an infinite scattered Hausdorff space. If K = I0K then K would be an infinite discrete space, hence not compact. Let x ∈ I1K and let H be a compact neighborhood of x such that H ∩ K 1 = {x} (recall that, as a compact Hausdorff space, K is locally compact). H must contain infinitely many points of I0K since otherwise x would be isolated in H and hence in K. Moreover, I0K is countable because it is a discrete subset of X. Thus H is a countable compact Hausdorff space consisting solely of isolated points except for a single limit point. Any such space is homeomorphic to ω + 1, contradicting Theorem 3.1. 2 Corollary 3.6. X is not locally compact. Lemma 3.7. Every paracompact space with a dense Lindelöf subspace is Lindelöf. Proof. See [8, p. 152].

2

Proposition 3.8. X is not paracompact. Proof. We have already shown that X is separable but is not Lindelöf; apply the above lemma. 2 We say that a subset A of X is almost dense if A is co-countable, or, equivalently, if there is some α < ω1 such that X α ⊆ A. Recall that a countably compact Hausdorff space is one in which every infinite subset has a limit point. Proposition 3.9. If X is countably compact then every infinite subset of X is almost dense. Proof. Suppose that X is countably compact and let A ⊆ X be infinite. A is a scattered space and thus has a relatively discrete subspace I0A . Furthermore, I0A is dense in A and I0A ⊆ A. Thus, without loss of generality, we may assume that A is relatively discrete. If A is not co-countable, Y = (X \ A) ∪ A ≈ X. But Y is not countably compact, since A has no limit points in Y . 2 Corollary 3.10. If X is countably compact then every open subset of X is either countable or co-finite. Proposition 3.11. If X is regular then X is not countably compact.

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Proof. Let x ∈ I1 and let U be any neighborhood of x that isolates x in X 1 . Using regularity, take V to be an open neighborhood of x such that V ⊆ U . Let Y = X \ {x}. Then Y ≈ X, but Y is not countably compact, since V ∩ Y is an infinite subset of Y with no limit points. 2 4. Filters on I0 We now take a closer look at the “bottom” of X. Specifically, we examine what convergence looks like in I0 ∪ I1 . Let U ⊆ X be a relatively discrete set and let x ∈ U . Define Fx (U ) = {V ∩ U : V is open and x ∈ V }. It is clear that • • • •

U If If U

/ F(U ). ∈ Fx (U ), ∅ ∈ A ∈ Fx (U ) and B ∈ Fx (U ) then A ∩ B ∈ Fx (U ). A ∈ Fx (U ) and A ⊆ B ⊆ U , B ∈ Fx (U ) (because U is discrete). \ {y} ∈ Fx (U ) for any x ∈ U (because X is T1 ).

In other words, Fx (U ) is a non-principal filter on the countable set U . In general, there is a duality between non-principal filters on ω and countable T1 topologies with a single non-isolated point. Let F be a filter on a set A and let G be a filter on a set B. We say that F and G are isomorphic (denoted F ≈ G) if there is a bijection f : A → B such that C∈F

⇔

f [C] ∈ G.

It is straightforward to show that, whenever U and V are neighborhoods of x and y, respectively, such that U ∩ I1 = {x} and V ∩ I1 = {y}, then Fx (U ∩ I0 ) ≈ Fy (V ∩ I0 ) if and only if U ≈ V . In other words, the duality between countably infinite Hausdorff spaces with a single non-isolated point and non-principal filters on ω extends to their respective isomorphisms. Recall that the Fréchet filter on an infinite set A is the set of all co-finite subsets of A. This filter is dual to the topology of ω + 1. By Theorem 3.1, there is no relatively discrete U ⊆ X and x ∈ U such that Fx (U ) is the Fréchet filter. If F is a filter on A and B ⊆ A, define F#B = {F ∩ B: F ∈ F} to be the restriction of F to B. It is clear that this set will be a filter if and only if B is in some ultrafilter extending F. We denote the set of all such B (alternatively, the union of all such ultrafilters) by F + : F + = {B: F#B is a filter}. A filter F is called homogeneous if F ≈ F#B whenever B ∈ F + . Lemma 4.1. Let F be a non-principal filter on a countably infinite set A and let B ∈ F. If F#B is not the Fréchet filter, then F ≈ F#B. Proof. Since F#B is neither the Fréchet filter nor a principal filter, there is some F ∈ F such that |B \ F | = ℵ0 . Since B \ F ⊆ A \ (B ∩ F ), we also have |A \ (B ∩ F )| = ℵ0 . Thus there is a bijection f : A \ (B ∩ F ) → B \ F . Define φ : A → B by

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 φ(x) =

f (x) if x ∈ A \ (B ∩ F ), x if x ∈ B ∩ F.

If C ⊆ A then C∈F

⇔

C ∩ B ∩ F = φ[C ∩ B ∩ F ] ∈ F

⇔

φ[C ∩ B ∩ F ] = φ[C] ∩ B ∩ F ∈ F#B

⇔

φ[C] ∈ F#B.

Hence φ is a filter isomorphism. 2 Theorem 4.2. Let U and V be open subsets of X such that |U ∩ X 1 | = |V ∩ X 1 | = 1. Then U ≈ V . Proof. It suffices to show that Fx (U ) ≈ Fy (V ) whenever U ∩ X 1 = {x} and V ∩ X 1 = {y}. First we will show that Fx (I0 ) = Fy (I0 ) for every x, y ∈ I1 . In order to prove this, pick x ∈ I1 and set   A = y ∈ I1 : Fx (I0 ) ≈ Fy (I0 ) . If A were countable, Y = X \ A would have the property that for no y ∈ I1Y = I1X ∩ Y is FyY (I0 ) isomorphic to FxX (I0 ); as FyY (I0 ) ≈ FyX (I0 ) and X ≈ Y , this is a contradiction. Thus A is uncountable and X ≈ I0 ∪ A. But it is clear that I1I0 ∪A = A, so every y ∈ I1I0 ∪A satisfies Fy (I0 ) ≈ Fx (I0 ). Hence all the Fx (I0 ) are isomorphic. Now let U , V be as above, and let x denote the unique element of U ∩ I1 and let y denote the unique element of V ∩ I1 . By Theorem 3.1, neither U nor V is homeomorphic to ω + 1. In the language of filters, this means that neither F(U ) nor F(V ) is the Fréchet filter. Furthermore, since X is T1 , neither Fx (U ∩ I0 ) nor Fy (V ∩ I0 ) is a principal filter. Thus, using the previous paragraph and Lemma 4.1, we have Fx (U ∩ I0 ) = Fx (I0 )#(U ∩ I0 ) ≈ Fx (I0 ) ≈ Fy (I0 ) ≈ Fy (I0 )#(V ∩ I0 ) = Fy (V ∩ I0 ).

2

Thus there is a single distinguished filter (up to isomorphism) given by X describing how I0 converges to each x ∈ I1 ; henceforth this filter will be denoted by F. F also tells us something about convergence in higher levels: Corollary 4.3. Let A be any countable relatively discrete subset of X and let x ∈ A. Then Fx (A) ≈ F. Proof. Let α = sup{rk(y) + 1: y ∈ A or y = x} and consider Y = A ∪ {x} ∪ X α . In Y , it is clear that A ∪ {x} is an open set and that (A ∪ {x}) ∩ Y 1 = {x}. It follows from Theorem 4.2 and the fact that X ≈ Y that Fx (A) ≈ F. 2 Corollary 4.4. F is homogeneous, does not have a countable base, is not the Fréchet filter, and is not principal. Proof. That F is homogeneous follows from the previous corollary, and that F does not have a countable base follows from the fact that no limit point of X has a countable neighborhood basis (by Corollary 3.2). That F is neither the Fréchet filter nor a principal filter follows from the fact that it does not have a countable base. 2 Theorem 4.5. F is not an ultrafilter.

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Proof. The following definition is due to Frolík (see [2]): if {Gn : n ∈ ω} is a collection of ultrafilters on ω   and if G is another ultrafilter on ω, G {Gn : n ∈ ω} consists of all sets of the form {Mn : n ∈ X} where X ∈ G and Mn ∈ Gn for each n ∈ X.  Assume that F is an ultrafilter; we show that F = G {Gn : n ∈ ω} where G ≈ G1 ≈ G2 ≈ · · · ≈ F, with all the Gn distinct. By a theorem of Frolík (see [1]), no such ultrafilter exists, so this is enough to prove the theorem. (In fact, a contradiction to Frolík’s theorem does not even require G1 ≈ G2 ≈ · · · ≈ F; it is enough to show merely that G ≈ F.) It should be acknowledge that this is essentially the idea used to prove Theorem 5.1(b) in [4]. Let x ∈ I2 and let U be an open neighborhood of x such that U ∩ X 2 = {x}. Let U ∩ I1 = {xn : n ∈ ω} and let Un be a neighborhood of xn such that Un ⊆ U and Un ∩X 1 = {xn }. Let Z = U ∩I0 . Let H = Fx (Z), let Gn = Fxn (Z), and let G be the filter defined on ω as follows: A ∈ G if and only if there is some open neighborhood V of x such that V ⊆ U and V ∩ I1 = {xn : n ∈ A}. In other words, G is the induced image of Fx ({xn : n ∈ ω}) under the natural isomorphism {xn : n ∈ ω} → ω. Let A ⊆ Z be an element of H. By definition, there is some open subset V of U such that x ∈ V and V ∩ Z = A. Since V is open, {n: xn ∈ V } ∈ G and, for each n ∈ G, V is a neighborhood of xn and   so A = V ∩ Z ∈ Fxn (Z). Thus A ∈ G {Gn : n ∈ ω}. As A was arbitrary, H ⊆ G {Gn : n ∈ ω}. By Corollary 4.3, H ≈ G ≈ G1 ≈ G2 ≈ · · · ≈ F. Since F (hence G, Gn ) is an ultrafilter, G {Gn : n ∈ ω} is also   an ultrafilter. Hence H ⊆ G {Gn : n ∈ ω} implies H = G {Gn : n ∈ ω}, which proves the theorem. 2 Given a filter G on a set A, define the filter G 2 on A × A by B ∈ G2

⇔

    n: m: (m, n) ∈ B ∈ G ∈ G.

If G ≈ G 2 , we say that G is idempotent. If there is a filter H on A × A such that H ⊆ G 2 and G ≈ H, then we say that G is sub-idempotent. In the original statement of the Toronto problem (in [6]) it was asserted that the existence of a nondiscrete HATS implies the existence of a homogeneous idempotent filter on ω, the implication being that the characteristic filter F described above is idempotent. This has not been proved, however, and the work of Gruenhage and Moore leads us to believe that it might be incorrect, as explained in the next paragraph. Of course we cannot say definitively that it is incorrect until a non-discrete HATS has been constructed. In [6], Stepr¯ ans defines an α-Toronto space, α < ω1 , to be a scattered space of height α that is homeomorphic to any of its subspaces of height α. The analogy with HATS is clear and, in fact, with this terminology a non-discrete HATS is just a thin–tall ω1 -Toronto space. Gruenhage and Moore, in [4], prove that α-Toronto spaces exist for α  ω and exist consistently for ω < α < ω1 . Their examples of α-Toronto spaces also satisfy Theorem 4.2: convergence in the bottom levels depends on a characteristic filter. However, the characteristic filters of the spaces they construct are not easily seen to be idempotent; indeed, they may not be. As we have no compelling reason to believe that the bottom few levels of a Toronto space should necessarily look very different from the bottom few levels of Gruenhage and Moore’s α-Toronto spaces, this leads us to believe that F need not be idempotent either. In the context of Gruenhage and Moore’s paper, there is a natural map taking their characteristic filter G to G 2 . This map does not prove that G is idempotent, but it does witness the fact that G is sub-idempotent (the proof of this is essentially identical to that of Proposition 4.6 below). In the case that X is regular, we can extend this result to non-discrete HATS: Proposition 4.6. If X is regular then F is sub-idempotent. Proof. Assume X is regular. We first observe that every x ∈ I1 has a neighborhood basis of clopen sets. Indeed, if we take any neighborhood U of x such that U ∩ X 1 = {x} and then take an open V ⊆ U with

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V ⊆ U , V must be clopen. This is because V \ V ⊆ I0 , and as every subset of I0 is open, this implies V = V . Furthermore, since x is the only non-isolated point of V , any neighborhood of x that is a subset of V is also clopen. Let x ∈ I2 and let U be a neighborhood of x such that U ∩ X 2 = {x}. Let V be a neighborhood of x such that V ⊆ U . Write I1 ∩ V = {yn : n ∈ ω}. We can choose a clopen neighborhood W0 of y0 isolating y0 in X 1 . Proceeding inductively, we can choose Wn+1 to be a clopen neighborhood of yn+1 isolating yn+1 in X 1 and disjoint from W1 , . . . , Wn . Furthermore, because Wn \ {yn } ⊆ I0 , every open subset of Wn that contains yn is also clopen. In particular, we may assume each Wn ⊆ V .  Let W = n∈ω Wn ∪ {x}; we would like for W to be an open neighborhood of x, but this may not be the case. Nonetheless, if we take Y = X \ (U \ W ), then Y ≈ X, x ∈ I2Y , and (in Y ) W is a clopen neighborhood of x. Thus, without loss of generality, we may assume that W is clopen. We argue that the filter    G = W  ∩ W ∩ I0  W  is a neighborhood of x is isomorphic to F and is weaker than a filter isomorphic to F 2 . It follows from this that F is sub-idempotent. Since x is a limit point of W ∩ I0 , Corollary 4.3 implies G ≈ F. To see that G is (isomorphic to a filter that is) weaker than F 2 , note that W ∩ I0 is partitioned into countably many countable subsets, namely the Wn ∩ I0 . We may think of these sets as the different “rows” of (W ∩ I0 ) × (W ∩ I0 ). Let Fn = Fyn (Wn ∩ I0 ) and H = Fx ({yn : n ∈ ω}). By Corollary 4.3, H ≈ F1 ≈ F2 ≈ · · · ≈ F. If A ∈ G then there is some neighborhood W  ⊆ W of x such that W  ∩ I0 = A. But then, because W  is open, A1 = {yn : yn ∈ W  } ∈ H and, for each yn ∈ A1 , {z ∈ Wn : z ∈ W } ∈ Fn (this is because W ∩ W  is a neighborhood of yn ). Thus A∈G

⇒

{n: A ∩ Wn ∈ Fn } ∈ H.

Since H ≈ F1 ≈ F2 ≈ · · · ≈ F, we conclude that G is weaker than F 2 .

2

Notice that this proposition does not rule out the idempotence of F: “sub-idempotent” does not necessarily mean strictly sub-idempotent. 5. Irregularity We say that x is a regular point of X if every open neighborhood of x contains a closed neighborhood of x; otherwise x is called an irregular point. X is a regular space if and only if all of its points are regular points. x ∈ X is a badly irregular point if every neighborhood of x is almost dense. We say that x is a barely  irregular point of X if x is irregular but has a neighborhood whose closure is contained in αrk(x) Iα . The main result of this section is: Theorem 5.1. If X is a non-discrete HATS then one of the following must hold: (1) X is regular. (2) There is an infinite closed R ⊆ I1 such that every x ∈ R is a regular point of X, and every other limit point of X is barely irregular. (3) Every limit point of X is badly irregular. Proof. We will prove Theorem 5.1 through a sequence of lemmas. Lemma 5.2. If Y is any topological space and x ∈ Z ⊆ Y , and if x is a regular point of Y , then x is a regular point of Z. In other words, regularity at a point is a hereditary property.

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Lemma 5.3. If Y is any topological space and D is dense in Y , then D ∩ U is dense in U for any open subset U of Y . Lemma 5.4. If X contains an irregular point that is not barely irregular, then every limit point of X is badly irregular. Proof. Assume that X contains an irregular point x that is not barely irregular. Let α = rk(x) > 0 and  consider Y = X \ 1β<α Iβ ≈ X. Because I0X ⊆ Y and I0X is dense in X, it follows from Lemma 5.3 that, for any open neighborhood U of x in Y , U Y = U X ∩ Y . Hence x is still an irregular point in Y and that x is still not barely irregular in Y . Furthermore, it is clear that rkY (x) = 1. Thus, if X contains an irregular point that is not barely irregular, it contains such a point in I1 . Within the scope of this proof, we will say that a point x of X is moderately irregular if every neighborhood of x has uncountable closure. (After this proof we will have no need of this new notion since, as we shall see shortly, every irregular point of X is either barely irregular or badly irregular.) Next we show that some point of I1 is moderately irregular. Suppose that this is not the case. Then every  point x of I1 has a neighborhood Ux such that Ux is countable. Set Y = (X \ x∈I1 Ux ) ∪ I0 ∪ I1 . Then X ≈ Y and every point of I1Y = I1X is barely irregular. This contradicts what we have shown so far, so I1 contains moderately irregular points. Next we show that every point of I1 is moderately irregular. Let A denote the set of moderately irregular points of I1 . If A is not co-countable, then Y = X \ A ≈ X has no moderately irregular points in I1Y , and we have already shown that this is impossible. Thus A is co-countable. Set Y = I0 ∪ A. Because X \ Y is countable and because I0 ⊆ Y , it follows from Lemma 5.3 that every moderately irregular point of X remains moderately irregular in Y . Hence every point of I1Y is moderately irregular in Y . As Y ≈ X, every point of I1X is moderately irregular in X. In particular, I1 contains no barely irregular points. Finally, we show that every limit point of X is badly irregular. Suppose this is not the case. Then, for some x ∈ Iα , α  1, there is a neighborhood U of x that is not almost dense, i.e., such that U is not co-countable. Set Y = (X \ U ) ∪ I0 ∪ {x}. Since U ∩ Y is a neighborhood of X in Y containing only x and points of I0Y , rkY (x) = 1. U ∩ Y Y = U ∩ Y ⊆ I0Y ∪ {x}, so x is not moderately irregular in Y . As Y ≈ X, this contradicts the conclusion of the previous paragraph. Thus every limit point of X is badly irregular. 2 Lemma 5.5. If x ∈ I1 then x is a regular point if and only if x has a clopen neighborhood U such that U ∩ X 1 = {x}. Moreover, this holds if and only if x has a neighborhood basis of clopen sets. Proof. If x ∈ I1 , there is an open set V such that V ∩ X 1 = {x}. Using the regularity of x, let U ⊆ V be an open neighborhood of x such that U ⊆ V . Then U \ U ⊆ I0 , and every subset of I0 is open; thus U = U , i.e., U is clopen. Conversely, suppose U is clopen and U ∩ I1 = {x}. Any open subset of U containing x is also clopen, so x has a neighborhood basis of clopen sets. 2 Lemma 5.6. Suppose that X is not regular and that not every limit point of X is badly irregular. Then X has countably many regular points, and all other points of X are barely irregular. Furthermore, the closure of the set of regular limit points of X is countable, and these points are arranged in X as follows: there is some 1 < α < ω1 such that there are no regular points of rank α or higher, and, for every β < α, there are infinitely many regular points of rank β. Proof. If X is not regular and not every limit point of X is badly irregular, then by Lemma 5.4 every limit point of X is either regular or barely irregular. Regularity at a point is a hereditary property, so

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the collection of regular points of X is a regular subspace of X. If this collection is uncountable then it is homeomorphic to X, making X regular. Thus X has countably many regular points. Let   α = sup rk(x) + 1: x is a regular point of X < ω1 . Clearly X has no regular points of rank α or higher. Next we show that α > 1, i.e., that X has regular limit points. We know that X has barely irregular points: let x be one such and let U be a neighborhood of X such that U is countable. Set Y = (X \U )∪(U ∩I0 )∪{x}. / I0 is a limit point of Y . U ∩ Y is a neighborhood of x in Y that contains only I0X = I0Y is dense in Y , so x ∈ x and points of I0Y , so rkY (x) = 1. Furthermore, U ∩ Y Y = U X ∩ Y = U , so U ∩ Y is a clopen neighborhood of x in Y . It follows from Lemma 5.5 that x is a regular limit point of Y . Since Y ≈ X, X has regular limit points. Let R denote the set of regular limit points of X. It remains to show that |R| = ℵ0 and that, for every 1  β < α, |R ∩ Iβ | = ℵ0 . Let β < α and let Rβ = R ∩ Iβ denote the set of regular points of rank β. If Rβ is finite, Y = X \ Rβ has no regular points of rank β; as Y ∼ = X, Rβ = ∅. Thus Rβ is either empty or infinite. Since β < α, there is  some γ such that β  γ < α and Rγ = ∅. Let Y = ζ<β Iζ ∪ X γ . If a point has rank γ in X then that same point has rank β in Y . Since the property of being a regular point is a hereditary property (Lemma 5.2), Y has regular points of rank β. As Y ≈ X, Rβ = ∅. Since, as we have already argued, Rβ cannot be finite and nonempty, |Rβ | = ℵ0 , and this holds for arbitrary 1  β < α. In order to show that R is countable, we must first show that some point of I1 is irregular in X. Let x be any irregular point of X. There is an open neighborhood U of x such that, for any neighborhood V of x, V  U . Consider Y = (X \ U ) ∪ (I0 ∩ U ) ∪ {x}. Clearly Y ≈ X and rkY (x) = 1, so it suffices to show that x is irregular in Y . Let V ∩ Y be an arbitrary neighborhood of x in Y . By Lemma 5.3 and the fact that I0X = I0Y is dense in both X and Y , V ∩ Y Y = V X ∩ Y . Hence



V ∩ Y Y \ (U ∩ Y ) = V X ∩ Y \ (U ∩ Y ) = V X \ U ∩ Y = V X \ U = ∅. Since V ∩ Y was arbitrary, U ∩ Y bears witness to the fact that x is not a regular point in Y . Thus X has irregular points in I1 . Suppose R is uncountable. Setting Y = R ∪ I0 , we have I1Y ⊆ R. It follows from Lemma 5.2 that every point of I1Y is a regular point of Y . Since Y ≈ X, this contradicts the conclusion of the previous paragraph, and it follows that R is countable. 2 All that remains for the proof of Theorem 5.1 is to show that in case (2) the set of regular points of X is a closed subset of I1 . For the remainder of the proof, we assume that X is not regular and does not have badly irregular points (i.e., that neither case (1) nor case (3) holds). As above, we use R to denote the set of regular points of X, α to denote the least ordinal such that R ∩ Iα = ∅, and Rβ to denote R ∩ Iβ . Lemma 5.7. Each Rβ , 0 < β < α, is closed in X. In particular, as R1 ⊆ I1 , R1 is closed and relatively discrete. Proof. R1 is countable because R is countable. Let Y = X \ (R1 \ R1 ) ≈ X. We claim that R1Y = R1X . R1X ⊆ R1Y because regularity at a point is a hereditary property. Suppose x ∈ R1Y \ R1X . Since X \ Y ⊆ X 2 , I0X = I0Y and I1X = I1Y . Thus x ∈ I1X . As x is barely irregular in X, x has a neighborhood U in X with U X ⊆ I0Y ∪ I1Y . As x ∈ R1Y , x has a clopen neighborhood V ⊆ I0Y ∪ {x} in Y . As V ⊆ I0X ∪ I1X and I0Y ∪ I1Y = I0X ∪ I1X is open in X, we have V ⊆ X, V is open in X, and U ∩ V is a neighborhood of x in X. Furthermore,

U ∩ V X ⊆ U X ∩ V X ⊆ U X ∩ V Y ∪ (Y \ X) ⊆ V Y = V.

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Since V ∩ I1X = {x}, U ∩ V X ⊆ I0 ∪ {x}. Since every subset of I0 is open, U ∩ V X is open, hence clopen, in X. It follows from Lemma 5.5 that x is a regular point of X. Thus x ∈ R1X and R1X = R1Y . R1Y Y = R1X Y = R1X X ∩ Y = R1X = R1Y . Thus R1Y is closed in Y and, as Y ≈ X, R1 is closed in X.  Now let 0 < β < α and let Y = X \ 0<γ<β Iγ . Y ≈ X and RβX = R1Y . Thus RβX is closed in Y . Since Y is the union of a closed subset of X (namely X β ) and I0 , and since Rβ ∩ I0 = ∅, Rβ is closed in X. 2 Lemma 5.8. α = 2, i.e., R ⊆ I1 . Proof. Suppose α > 2 and let X0 = X. Take X1 = (X0 \ I1X0 ) ∪ R1X0 . Clearly X1 is co-countable in X. R1X0 ⊆ R1X1 : this is because regularity is hereditary, R1X0 ⊆ X1 , and, as I0X0 ⊆ X1 , every rank-1 point of X0 remains a rank-1 point in X1 . Also, R1X0 = R1X1 : this is because R1 is closed, which implies that every x ∈ R2X0 becomes a rank-1 point in X1 and, as regularity is hereditary, a member of R1X1 . Using transfinite recursion, we now define a sequence Xβ : β < ω1  of subsets of X. Let X0 = X. Given  X X Xβ , we let Xβ+1 = (Xβ \ I1 β ) ∪ R1 β . For limit β, we take Xβ = γ<β Xγ . It follows from induction and the argument of the previous paragraph that, for every β < ω1 , Xβ is co-countable in X, hence homeomorphic X X X to X, and R1 β ⊆ R1 β+1 = R1 β .  X X X +1 Take Y = β<ω1 R1 β . Because R1 β  R1 β for every β, Y is uncountable and hence homeomorphic   X X Y to X. There is some γ < ω1 such that I0 ⊆ β<γ R1 β . Thus β<γ+1 R1 β is not relatively discrete. On   X X X the other hand, β<γ+1 R1 β ⊆ R1 γ+1 , so, by Lemma 5.7 and the fact that Xγ+1 ≈ X, β<γ+1 R1 β is relatively discrete. This is a contradiction, so α = 2. 2 This completes the proof of Theorem 5.1.

2

The method of the proof of Lemma 5.8 can be abstracted into a more general result: Proposition 5.9. Assume X satisfies (2) from Theorem 5.1. If D is a countable subset of X containing no regular points, then RX = RX\D . Proof. Stretching our notation slightly, if Y ⊆ X and h : X → Y is a homeomorphism, we take DY to mean h[D]. When a homeomorphism h is not mentioned explicitly, it is assumed that h has been chosen arbitrarily and fixed. Assume that D contains no regular points and that RX = RX\D . Because every isolated point is a regular point, D ∩ I0 = ∅, so if rkX (x) = 1 then rkX\D (x) = 1. Since regularity is hereditary, this implies RX ⊆ RX\D . Thus RX  RX\D . As in the proof of Lemma 5.8, we define a transfinite sequence of  uncountable subsets of X by taking X0 = X, Xβ+1 = Xβ \ DXβ , and, for limit α, Xα = β<α Xβ . For each β < ω1 , Xβ ≈ X and RXβ  RXβ+1 .  Let Y = β<ω1 RXβ . Since RXβ  RXβ +1 for all β < ω1 , Y is uncountable and Y ≈ X. As in the proof   of Lemma 5.8, there is some γ < ω1 such that I0Y ⊆ β<γ RXβ . Thus β<γ+1 RXβ is not relatively discrete.   X On the other hand, β<γ+1 RXβ ⊆ RXγ+1 , so, by Lemma 5.7 and the fact that Xγ+1 ≈ X, β<γ+1 R1 β is relatively discrete, a contradiction. 2 Corollary 5.10. Assume X satisfies (2) from Theorem 5.1. If U is an open subset of X such that R ⊆ U then X 1 ⊆ U . In other words, it is impossible to separate any limit point of X from R by open sets. 1

Proof. Let U be open, R ⊆ U , and suppose there is some x ∈ X 1 \ U . Because I1X = I0X is dense in X 1 , we may assume x ∈ I1 . Let V be an open neighborhood of x such that V ⊆ X 1 \ U , such that V ⊆ {x} ∪ I0 , and such that V ⊆ I0 ∪ I1 (for this last requirement, recall that x is barely irregular). Clearly V ∩ U = ∅ so, in particular, V ∩ R = ∅. Let ∂V = V \ V . By Proposition 5.9, RX = RX\∂V ; thus x is irregular in X \ ∂V .

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On the other hand, V is a clopen neighborhood of x in X \ ∂V , and it follows from Lemma 5.5 that x is regular in X \ ∂V . 2 6. Toronto spaces that are not Hausdorff We now abandon HATS and consider Toronto spaces that are not Hausdorff. The lower topology on a totally ordered set X is the topology with sets of the form (−∞, x] as a basis, and the upper topology on X is the topology with sets of the form [x, ∞) as a basis. Alternatively, a set is open in the lower topology if and only if it is an initial segment, and a set is open in the upper topology if and only if it is a final segment. Notice that, if κ is a cardinal, the lower and upper topologies on κ are examples of Toronto spaces. This is because every full-cardinality subset Y of κ has the same order type as κ; since a subset of Y will be open in Y if and only if it is an initial or a final segment, respectively, the natural order-isomorphism from Y to κ is a homeomorphism. The next theorem tells us that, other than the trivial topology on κ points, these are the only Toronto spaces of size κ that are not T1 . Theorem 6.1. Let κ be an infinite cardinal. Up to homeomorphism, there are exactly three Toronto spaces of size κ that are not T1 : the indiscrete topology and the lower and upper topologies on κ. Proof. Let X be a Toronto space of size κ that is not T1 . We distinguish three cases: either X is indiscrete, X has an open set of size κ other than X, or X has a closed set of size κ other than X. It is easy to see that these three cases are exhaustive. We will show that in the second case X is homeomorphic to the upper topology on κ, and in the third case X is homeomorphic to the lower topology on κ. Consider the case that X has a nontrivial open subset of size κ. We prove first the following claim: there is some x0 ∈ X such that {x0 } = {x0 } and, for any x ∈ X, x0 ∈ {x}. Let U be an open subset of X of size κ, x ∈ X \ U , and set Y = U ∪ {x}. X ≈ Y and Y contains a closed singleton, namely {x}. Let C denote the set of closed singletons in X. We have just shown that C = ∅. Also, |C| < κ: if |C| = κ then X ≈ C and, as every singleton in C is closed, X is T1 , contrary to assumption. ˜ Set Y = X \ C˜ and suppose that |Y | = κ. Since Let C˜ = {x ∈ X: {x} ∩ C = ∅}; we claim that X = C. X is a Toronto space, Y ≈ X and there is a nonempty set C Y of closed singletons in Y . Let y ∈ C Y . Since y∈ / C X and {y}Y = Y ∩ {y}X , we must have {y}X ∩ C˜ X = ∅. But the relation y ∈ {x} is transitive, and it follows from the definition of C˜ that {y}X ∩ C X = ∅. Thus y ∈ C˜ X , a contradiction. Thus |Y | < κ and, ˜ = κ. It is straightforward to show that C X = C C˜ and that C˜ X = C˜ C˜ . Since X ≈ C˜ and consequently, |C| ˜ ˜ we have X = C˜ X . C˜ C = C, For each x ∈ C, let Kx = {y ∈ X: x ∈ {y}}. Suppose there is some x0 ∈ C such that |Kx0 | = κ. Then X ≈ Kx0 and, for every point y ∈ Kx0 , x0 ∈ {y}Kx0 ; in this case the claim is proved. Now suppose that,  for every x ∈ C, |Kx | < κ (note that, since |C| < κ and X = C˜ = x∈C Kx , this can only happen for singular κ). Let x ∈ C and consider Y = (X \ Kx ) ∪ {x}. Y ≈ X and it is clear that KxY = {x}. Setting C  = {x ∈ C: Kx = {x}}, this shows that C  = ∅. Setting Y = X \ C  , we have Y ≈ X and C  Y = ∅. This contradiction establishes that for some (every) x0 ∈ C, |Kx0 | = κ. As above, then, we have C = {x0 } and X = {y ∈ X: x0 ∈ {y}}. This finishes the proof of our claim. We now will use transfinite recursion to find a sequence xα : α < κ of points in X such that {xα+1 : α < κ} is homeomorphic to the upper topology on κ. Let x0 be the unique point of X such that {x0 } = {x0 }. Assuming xβ : β < α has already been defined (for α a successor or a limit), take xα to be the unique point of Xα = X \ {xβ : β < α} such that {xα }Xα = {xα }. Let Y = {xα : α < κ}. It follows from a straightforward transfinite induction that {xα }Y = {xβ : β  α} for every α < κ. Taking Z = {xα+1 : α < κ}, we may conclude that A is a closed subset of Z if and only if A is an initial segment of Z (i.e., A = {xβ : β < α} for some α  κ). Thus Z is homeomorphic to the upper topology on κ, and X ≈ Z.

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Now suppose that X has a nontrivial closed subset of size κ. Whereas before X contained a closed singleton, now X must contain an open singleton. For x ∈ X, define x◦ to be the intersection of all open sets containing x; this mimics the idea of {x} for open sets, except that x◦ is not in general open. As before, set C = {x ∈ X: x◦ = {x}}. C = ∅ because X contains an open singleton and, as a subspace of X, C carries the discrete topology, so |C| < κ. Arguing as before, it is possible to show that there is a unique x0 ∈ X such that x◦0 = {x0 } and, for every y ∈ X, x0 ∈ y ◦ . Also, {x0 } must be open in X because X must contain an open singleton. Using transfinite recursion as before, we obtain a subset Y = {xα : α < κ} of X such that A ⊆ Y is open if and only if A is an initial segment of Y . Thus X is homeomorphic to the lower topology on κ. 2 It should be noted that the following corollary to Theorem 6.1 is asserted without proof in [7], and should be considered another of the many “folklore” results surrounding Toronto spaces. Corollary 6.2. Up to homeomorphism, there are exactly five Toronto spaces of size ℵ0 : the discrete topology, the indiscrete topology, the co-finite topology, and the lower and upper topologies on ω. Proof. Let X be a countable Toronto space and suppose that X is T1 , i.e., that every co-finite set is open in X. If X has also some open co-infinite subset A then, taking some x ∈ A, Y = {x} ∪ X \ A is an infinite subspace of X that contains a clopen singleton {x}. Since Y ≈ X, X contains a clopen singleton. Let I denote the set of clopen singletons in X. If I is finite then X \ I is a space without clopen singletons that is homeomorphic to X, a contradiction. If I is infinite, X ≈ I and X is discrete. Thus if X is T1 then X has either the co-finite topology or the discrete topology, and the result follows from Theorem 6.1. 2 In order to state the next result succinctly, let LT be the lower topology on ω1 , let UT be the upper topology on ω1 , let CF be the topology of co-finite subsets of ω1 , and let CC be the topology of co-countable subsets of ω1 . If σ and τ are two different topologies on some set X, let σ, τ  denote the topology on X generated by σ ∪ τ . Proposition 6.3. There are at least eight Toronto spaces of size ℵ1 : the discrete and indiscrete topologies, LT , UT , CF , CC , LT , CF , and UT , CF . Any HATS is a refinement of LT , CF , and any other Toronto space of size ℵ1 either is a refinement of LT , CF  or is strictly between CF and CC . Proof. It is easy to check that each of these eight spaces is Toronto. Notice that LT , UT  and LT , CC  are both the discrete topology, and UT , CC  = CF , CC  = CC . Thus any subcollection of these eight topologies generates one of these eight topologies. We have seen that if X is a HATS then X is a scattered space with height ω1 and width ω. Thus we may assume that X = ω1 and that X α = ω1 \ ω · α. It is then easy to see that every initial segment of ω1 is open and, since X is Hausdorff, so is every co-finite set. Hence X refines LT , CF . Examining the proof of Proposition 2.2, we see that a non-discrete Toronto space X is a scattered space of height ω1 and width ω if (and only if) X contains a clopen singleton. Then, by the argument of the previous paragraph, X is a refinement of LT , CF . Thus, to finish the proof, it suffices to show that any Toronto space not on our list either contains a clopen singleton or is strictly between CF and CC . Suppose X is a Toronto space other than the eight listed above. By Theorem 6.1, X refines CF . If X is not strictly coarser than CC , X contains a nonempty open set A with uncountable complement. Taking x ∈ A and Y = {x} ∪ (X \ A), {x} is a clopen singleton in Y and Y ≈ X. 2 References [1] Z. Frolík, Fixed point maps of βN, Bull. Am. Math. Soc. 74 (1968) 187–191. [2] Z. Frolík, Sums of ultrafilters, Bull. Am. Math. Soc. 73 (1) (1967) 87–91.

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[3] C. Good, K. Papadopoulos, A topological characterization of ordinals: van Dalen and Wattel revisited, Topol. Appl. 159 (2012) 1565–1572. [4] G. Gruenhage, J.T. Moore, Countable Toronto spaces, Fundam. Math. 163 (2) (2000) 143–162. [5] M. Rajagopalan, A chain compact space which is not strongly scattered, Isr. J. Math. 23 (2) (1976) 117–125. [6] J. Steprans, Stepr¯ ans’ problems, in: J. van Mill, G.M. Reed (Eds.), Open Problems in Topology, North-Holland, Amsterdam, 1990, pp. 13–20. [7] Toronto space, Wikipedia: The Free Encyclopedia, Wikimedia Foundation, Inc., 2004: http://en.wikipedia.org/wiki/ Toronto_space. [8] S. Willard, General Topology, Dover Publications, Inc., Mineola, New York, 1970.